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3 years ago

What is the specific heat of the solid phase? (Please see picture attached)

Chemistry

1 answer:

ValentinkaMS [17]3 years ago

4 0

Answer:

B.0.2 J/g°C

Explanation:

From the attached picture;

Heat attained in the solid phase is 200 Joules
Change in temperature is 50°C ( from 0°C to 50°C)
Mass of the solid is 20 g

We are required to determine the specific heat capacity of the substance;

We need to know that Quantity of heat is given by the product of mass,specific heat capacity and change in temperature.
That is; Q = mcΔT

Rearranging the formula;

c = Q ÷ mΔT

Therefore;

Specific heat = 200 J ÷ (20 g × 50°c)

= 0.2 J/g°C

Thus, the specific heat of the solid is 0.2 J/g°C

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A gas mixture with a total pressure of 745 mmHg contains each of the following gases at the indicated partial pressures: CO2, 24

Setler79 [48]

Answer:

For Part A: The partial pressure of Helium is 218 mmHg.

For Part B: The mass of helium gas is 0.504 g.

Explanation:

For Part A:

We are given:

$p_{CO_2}=245mmHg\\p_Ar}=119mmHg\\p_{O_2}=163mmHg\\P=745mmHg$

To calculate the partial pressure of helium, we use the formula:

$P=p_{CO_2}+p_{Ar}+p_{O_2}+p_{He}$

Putting values in above equation, we get:

$745=245+119+163+p_{He}\\p_{He}=218mmHg$

Hence, the partial pressure of Helium is 218 mmHg.

For Part B:

To calculate the mass of helium gas, we use the equation given by ideal gas:

PV = nRT

or,

$PV=\frac{m}{M}RT$

where,

P = Pressure of helium gas = 218 mmHg

V = Volume of the helium gas = 10.2 L

m = Mass of helium gas = ? g

M = Molar mass of helium gas = 4 g/mol

R = Gas constant = $62.3637\text{ L.mmHg }mol^{-1}K^{-1}$

T = Temperature of helium gas = 283 K

Putting values in above equation, we get:

$218mmHg\times 10.2L=\frac{m}{4g/mol}\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 283K\\\\m=0.504g$

Hence, the mass of helium gas is 0.504 g.

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2:2 so the same proportion
0,03

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AveGali [126]

The answer to this question would be:
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One litre of hydrogen at STP weight 0.09gm of 2 litre of gas at STP weight 2.880gm. Calculate the vapour density and molecular w

expeople1 [14]

Answer:

we know, at STP ( standard temperature and pressure).

we know, volume of 1 mole of gas = 22.4L

weight of 1 Litre of hydrogen gas = 0.09g

so, weight of 22.4 litres of hydrogen gas = 22.4 × 0.09 = 2.016g ≈ 2g = molecular weight of hydrogen gas.

similarly,

weight of 2L of a gas = 2.88gm

so, weight of 22.4 L of the gas = 2.88 × 22.4/2 = 2.88 × 11.2 = 32.256g

hence, molecular weight of the gas = 32.256g

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hence, vapor density of the gas is 16.128g.

Explanation:

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